Prove that : \( \int_{0}^{\infty} \frac{x^{2}}{\left(x^{2}+1\right)^{2}} d x=\frac{\pi}{4} \). सिद्ध कीजिए : \( \int_{0}^{\infty} \frac{x^{2}}{\left(x^{2}+1\right)^{2}} d x=\frac{\pi}{4} \).

To prove that \[ \int_{0}^{\infty} \frac{x^{2}}{\left(x^{2}+1\right)^{2}} \, dx = \frac{\pi}{4}, \] we can evaluate the integral using a strategic decomposition and known integral results. ### Step 1: Simplify the Integrand First, observe that the integrand can be rewritten by splitting the fraction: \[ \frac{x^{2}}{\left(x^{2}+1\right)^{2}} = \frac{(x^{2} + 1) - 1}{\left(x^{2}+1\right)^{2}} = \frac{1}{x^{2} + 1} - \frac{1}{\left(x^{2}+1\right)^{2}}. \] So, the integral becomes: \[ \int_{0}^{\infty} \frac{x^{2}}{\left(x^{2}+1\right)^{2}} \, dx = \int_{0}^{\infty} \left( \frac{1}{x^{2} + 1} - \frac{1}{\left(x^{2}+1\right)^{2}} \right) dx. \] ### Step 2: Evaluate the Individual Integrals Now, we can evaluate each integral separately. 1. **First Integral:** \[ \int_{0}^{\infty} \frac{1}{x^{2} + 1} \, dx \] This is a standard integral that evaluates to: \[ \int_{0}^{\infty} \frac{1}{x^{2} + 1} \, dx = \frac{\pi}{2}. \] 2. **Second Integral:** \[ \int_{0}^{\infty} \frac{1}{\left(x^{2}+1\right)^{2}} \, dx \] This integral can be evaluated using a known result or by substitution. The standard result is: \[ \int_{0}^{\infty} \frac{1}{\left(x^{2}+1\right)^{2}} \, dx = \frac{\pi}{4}. \] ### Step 3: Combine the Results Subtract the second integral from the first: \[ \int_{0}^{\infty} \frac{x^{2}}{\left(x^{2}+1\right)^{2}} \, dx = \frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4}. \] ### Conclusion Thus, we have proven that: \[ \int_{0}^{\infty} \frac{x^{2}}{\left(x^{2}+1\right)^{2}} \, dx = \frac{\pi}{4}. \]